lecture 6 power functions

Solve $\sqrt{2x+5}>x+1$.

Best way is to graph the equations, and graphically solve for the solution range. That always works.

Not recommended to use below method.

Restrict $2x+5\ge 0$
$x\ge \frac{-5}{2}$

$\sqrt{2x+5}>x+1$

Maybe assume both sides are positive, and square both sides.

$2x+5>x^2+2x+1$
$0>x^2-4$
$0>(x+2)(x-2)$

This means $-2<x<2$

Therefore.

$x\in [-\frac{-5}{2},2)$

Graph $f(x)=\frac{6}{\sqrt{9-3x}}+1$.

Rewrite.
$f(x)=a[b(x-h){]}^n+k$

$f(x)=6[-3(x-3){]}^{\frac{-1}{2}}+1$

Recall the 9 classes of power functions table.

Step one find the asymptotes. They are $x=3$ and $y=1$

There is a reflection in y axis, as seen in $-3$.

Now we know how to graph it yay.

Find domain and range of $y=-(3-x{)}^{\frac{3}{4}}-1$.

Endpoint at $(3,-1)$.

We know the shape of a power function that is between 0 and 1 and is odd on even. This one is reflected in both the x axis and y axis.

The domain is $(-\infty ,3]$ and the range is $(-\infty ,-1]$.

Find the domain of $y=\sqrt{1-4x^2}$.

Restrict.

$1-4x^2\ge 0$
$4x^2\le 1$
$x^2\le \frac{1}{4}$
$-\frac{1}{2}\le x\le \frac{1}{2}$

Find domain and range of $y=\sqrt{2x^2-9x-5}$.

Restrict.

$2x^2-9x-5\ge 0$

We can graph the quadratic $2x^2-9x-5=0$ and find where the graph $\ge 0$.

Has positive shape.

$(2x+1)(x-5)=0$

X-intercepts $x=5$ and $x=\frac{-1}{2}$

Therefore the domain of y is $x\in (-\infty ,\frac{-1}{2}]\cup [5,\infty )$

Now find range of y. The things under square root always $\ge 0$, so the range is $[0,\infty )$.

Find asymptotes of the function $f(x)=\frac{x+4}{2x-3}$.

First we need to rearrange. Umm linear over linear is something used in this step.

$f(x)=\frac{1}{2}+\frac{11}{2(2x-3)}$

Therefore the asymptotes are $x=\frac{3}{2}$ and $y=\frac{1}{2}$

Let $f(x)=\frac{3}{(x+1{)}^3}-2$. Find asymptotes of $f^{-1}$.

We don’t even need to find inverse. The original asymptotes are $x=-1$ and $y=-2$, so the inverse has the asymptotes $x=-2$ and $y=-1$.

But to find f inverse just for practice…
$x=\frac{3}{(y+1{)}^3}-2$

$(y+1{)}^3=\frac{3}{x+2}$
$y=(\frac{3}{x+2}{)}^{\frac{1}{3}}-1$
$y=3^{\frac{1}{3}}(x+2{)}^{\frac{-1}{3}}-1$

The lines $y=-ax+1$ and $y=\frac{-x}{3}+2b$ are inverse of each other. Find values of a and b.

Equation one. $y=-ax+1$. The inverse of equation one is $y=\frac{x-1}{-a}$. And this is the same as $y=\frac{-x}{3}+2b$. We have an equation now.

$\frac{x-1}{-a}=\frac{-x}{3}+2b$
$\frac{-x+1}{a}=\frac{-x}{3}+2b$
$\frac{-x}{a}+\frac{1}{a}=\frac{-x}{3}+2b$

Equating coefficients.

$a=3$
$\frac{1}{a}=2b$
$b=\frac{1}{6}$

Let $f:(n,\infty )\to \mathbb{R},f(x)=\frac{m}{(x-n{)}^2}$ and $m.n\in {\mathbb{R}}^{+}$. Find the rule and domain and range of $f^{-1}$.

So function f is a hyperbola with the asymptotes $x=n$ and $y=0$ and the domain restriction $x>n$. The range is therefore $(0,\infty )$.

The inverse function will have asymptotes $x=0$ and $y=n$.

For inverse.

$x=\frac{m}{(y-n{)}^2}$
$y-n=\pm \sqrt{\frac{m}{x}}$
$f^{-1}(x)=n\pm \sqrt{\frac{m}{x}}$

From the domain and range of f, the inverse function has range $(n,\infty )$ and domain $(0,\infty )$.

Given $h(x)=\frac{4x}{5-x}$ and $f(x)=h^{-1}(x)$, find $f(2)$ and state whether f is increasing or decreasing at the point.

Rearrange with linear over linear.

$h(x)=-4+\frac{20}{5-x}$

Find f.

$x=-4+\frac{20}{5-y}$
$x+4=\frac{20}{5-y}$
$5-y=\frac{20}{x+4}$
$y-5=\frac{-20}{x+4}$
$y=\frac{-20}{x+4}+5$
$f(x)=\frac{-20}{x+4}+5$

$f(2)=\frac{5}{3}$

Graph is a hyperbola with middle point at $(-4,5)$ and reflected once. You can visualize this and find that the slope is always positive when $x>-4$. Therefore it is increasing at the point.

Given $h(x)=\frac{4x}{5-x}$ and $f(x)=h^{-1}(x)$, find $f(2)$ and state whether f is increasing or decreasing at the point.

$x=\frac{4y}{5-y}$
Rearrange.

$x=-4+\frac{20}{5-y}$
$x+4=\frac{20}{5-y}$
$5-y=\frac{20}{x+4}$
$y-5=\frac{-20}{x+4}$
$f(x)=-20(x+4{)}^{-1}+5$
$f(x)=\frac{5}{3}$

Visualize the graph. The graph has a positive slope when $x>-4$.

Given $h(x)=(2x+11{)}^{\frac{1}{5}}$ and $f(x)=h^{-1}(x)$, find $f(2)$ and state whether f is increasing or decreasing at the point.

$x=(2y+11{)}^{\frac{1}{5}}$
$f(x)=\frac{x^5-11}{2}$
$f(2)=\frac{21}{2}$

Try to visualize. The graph has a positive slope at the point.

Given $h(x)=(x-2{)}^3+1$ and $f(x)=h^{-1}(x)$, find $f(3)$ and state whether f is increasing or decreasing at the point.

$x=(y-2{)}^3+1$
$(x-1{)}^{\frac{1}{3}}=y-2$
$f(x)=(x-1{)}^{\frac{1}{3}}+2$

$f(3)=2^{\frac{1}{3}}+2$

The graph of f has a middle point $(1,2)$ and we know the shape of power function with exponent between 0 and 1 and is odd over odd. At $x>1$ the slope is positive.

Given $h(x)=\frac{1+2x}{x+7}$ and $f(x)=h^{-1}(x)$, find $f(3)$ and state whether f is increasing or decreasing at the point.

$x=\frac{1+2y}{y+7}$
$x=2+\frac{-13}{y+7}$
$y+7=\frac{-13}{x-2}$
$f(x)=\frac{-13}{x-2}-7$
$f(3)=-20$

Visualize the graph with the asymptotes. When $x>2$ the slope is positive.

Let $h:[0,\infty )\to \mathbb{R},h(x)=\frac{7}{x+2}-3$. Find $h^{-1}$.

$x=\frac{7}{y+2}-3$
$y+2=\frac{7}{x+3}$
$h^{-1}(x)=\frac{7}{x+3}-2$

Find asymptotes of $f(x)=\frac{4-x}{2-x}$.

Rearrange with linear over linear.

$f(x)=1+\frac{2}{2-x}$

Therefore asymptotes at $x=2$ and $y=1$.

Find asymptotes of $f(x)=\frac{x+2}{5-x}$.

Rearrange with linear over linear.

$f(x)=-1+\frac{7}{5-x}$

Therefore asymptotes at $x=5$ and $y=-1$.

Find $g(x)=f(f^{-1}(x)-1)$ where $f(x)=3e^{1-x}$.

$x=3e^{1-y}$
$\frac{x}{3}=e^{1-y}$
$\ln (\frac{x}{3})=1-y$
$f^{-1}(x)=-\ln (\frac{x}{3})+1$

$g(x)=3e^{1-[-\ln (\frac{x}{3})+1-1]}$
$g(x)=3e^{1+\ln (\frac{x}{3})}$

$g(x)=3e^{\ln (e)+\ln (\frac{x}{3})}$

$g(x)=3\cdot e^{\ln (e\cdot \frac{x}{3})}$

$g(x)=3\cdot e^{\ln (\frac{ex}{3})}$

$g(x)=3(\frac{ex}{3})$
$g(x)=ex$

Solve $\sqrt{2x-3}>x-3$. Without graphing.

We have a restriction $2x-3\ge 0$ which is $x\ge \frac{3}{2}$.

Need three cases. Left hand side always positive, but right hand side could be anything.

Case $x-3>0$.
$\sqrt{2x-3}>x-3$
$2x-3>(x-3{)}^2$
$0>x^2-6x+9-2x+3$
$0>x^2-8x+12$
$0>(x-6)(x-2)$

Therefore x must be in $(2,6)$.

Remember to satisfy the case, $x\ge 3$. And satisfy initial restriction too.

$x\in [3,6)$

Case $x-3=0$.
$\sqrt{2x-3}=x-3$
$0=(x-6)(x-2)$

$x=6\vee x=2$

Remember to satisfy the case, $x=3$. And satisfy initial restriction too.
No solutions for this case.

Case $x-3<0$.
All values of x will satisfy $\sqrt{2x-3}>x-3$.

Remember to satisfy the case, $x<3$. And satisfy initial restriction too.

$x\in [\frac{3}{2},3)$

Combine the three cases we found.

$x\in [\frac{3}{2},6)$

Given $f:[-1,2]\to \mathbb{R},f(x)=-4x^3$ is one-to-one, find the domain of $f^{-1}$.

Since we know it is just one-to-one, the range of $f(x)$ is $[-32,4]$.

The domain of the inverse is also $[-32,4]$.

Given $f:(1,2)\to \mathbb{R},f(x)=\frac{1}{2}{x}^4$ is one-to-one, find the domain of $f^{-1}$.

Since it is one-to-one, the range of f is $(\frac{1}{2},8)$.

The domain of the inverse is also $(\frac{1}{2},8$).

Find domain and range of $f(x)=2-\sqrt{5+8x}$.

Restriction is $5+8x\ge 0$ which is $x\ge \frac{-5}{8}$.

Domain is $[\frac{-5}{8},\infty )$.

The endpoint is $(\frac{-5}{8},2)$.

There is a reflection in x axis.

Therefore the range is $(-\infty ,2]$.

The function $f:(-\infty ,a]\to \mathbb{R},f(x)=\ln (x^4)$ will have an inverse for what values of a?

F has a restriction $x^4>0$, and it is always satisfied. So all the domain is all numbers.

We need to restrict to get half of the function, there is asymptote at $x=0$ so we want $a<0$

Let $f(x)=a{x}^3+b{x}^2+cx+d$ where a,b,c,d are positive. For what b will $f(x)$ have an inverse function?

For there to be an inverse function, the function needs to be one-to-one.

There can be at most one point where the derivative is zero.

$f^{\prime }(x)=3a{x}^2+2bx+c$

A stationary point will mean $3a{x}^2+2bx+c=0$. And we can have at most one stationary point.

So we want the discriminant of this to be zero or negative.

$\Delta =b^2-4ac=4b^2-4(3ac)=4b^2-12ac$
$4b^2-12ac=0$ or $4b^2-12ac<0$
$b^2=3ac$ or $b^2<3ac$

We know that b is positive. No need take absolute value.
$b=\sqrt{3ac}$ or $b<\sqrt{3ac}$

Remember original restriction b is positive.
$0<b\le \sqrt{3ac}$

Let $f(x)=a{x}^3+b{x}^2+cx+d$ where a,b,c,d are negative. For what b will $f(x)$ not have an inverse function?

The function needs to be not one-to-one.

There is more than one point where derivative is zero.

$f^{\prime }(x)=3a{x}^2+2bx+c$
$0=3a{x}^2+2bx+c$

Discriminant is positive.

$\Delta =4b^2-12ac$

$4b^2-12ac>0$
$b^2>3ac$
We know b is negative. Need to take absolute value.
$|b|>\sqrt{3ac}$
$b<-\sqrt{3ac}$ or $b>\sqrt{3ac}$
Remember original restriction b is negative.
$b<-\sqrt{3ac}$

Let $f(x)=x^3-2x^2+ax-1$ and a is positive. For what a will $f(x)$ have inverse?

For there to be an inverse function, the function needs to be one-to-one.

There can be at most one point where the derivative is zero.

$f^{\prime }(x)=3x^2-4x+a$
$0=3x^2-4x+a$

So we want the discriminant of this to be zero or negative.

$\Delta =16-4(3a)=16-12a$
$16-12a=0$ or $16-12a<0$
$a=\frac{4}{3}$ or $a>\frac{4}{3}$

$a\ge \frac{4}{3}$

Let $f(x)=a{x}^3+2x^2-x-1$ and a is positive. For what a will $f(x)$ have inverse?

For there to be an inverse function, the function needs to be one-to-one.

There can be at most one point where the derivative is zero.

$f^{\prime }(x)=3a{x}^2+4x-1$
$0=3a{x}^2+4x-1$

So we want the discriminant of this to be zero or negative.

$\Delta =16+12a$

$16+12a=0$ or $16+12a<0$
$a=\frac{-4}{3}$ or $a<\frac{-4}{3}$
Remember original restriction a is positive.
No solutions.

Let $f(x)=x^3+3a{x}^2+ax+1$. For what a will $f(x)$ not have inverse?

For there to be no inverse, the function is not one-to-one.

There is more than one point where derivative is zero.

$f^{\prime }(x)=3x^2+6ax+a$
$0=3x^2+6ax+a$

We want discriminant to be positive.
$\Delta =36a^2-4(3a)=36a^2-12a$
$36a^2-12a>0$
$3a^2-a>0$
$a(3a-1)>0$

Case $a>0$
$3a-1>0$
$a>\frac{1}{3}$
So $a>\frac{1}{3}$

Case $a=0$
$0>0$
Not possible.

Case $a<0$
$3a-1<0$
$a<\frac{1}{3}$
So $a<\frac{1}{3}$

$a>\frac{1}{3}$ or $a<\frac{1}{3}$

Let $f(x)=x^3-a{x}^2+3x-1$. For what a will $f(x)$ have inverse?

For there to be an inverse function, the function needs to be one-to-one.

For a cubic polynomial to be one-to-one, there can be at most one point where the derivative is zero.

$f^{\prime }(x)=3x^2-2ax+3$
$0=3x^2-2ax+3$

Discriminant must be zero or negative.

$\Delta =4a^2-4(9)=4a^2-36$
$4a^2-36=0$ or $4a^2-36<0$
$a^2=9$ or $a^2<9$
$a^2=9$ or $|a|<3$
$a^2=\pm 3$ or $-3<a<3$

$a\in [-3,3]$

For $f:(-\infty ,0)\to \mathbb{R},f(x)=3-\frac{4}{\sqrt{1-x}}$, what is the domain and rule for inverse?

$x=3-\frac{4}{\sqrt{1-y}}$
$3-x=\frac{4}{\sqrt{1-y}}$
$\sqrt{1-y}=\frac{4}{3-x}$

$1-y=\frac{16}{(3-x{)}^2}$

$y-1=\frac{-16}{(3-x{)}^2}$

$f^{-1}(x)=\frac{-16}{(3-x{)}^2}+1$

The range of original is $(-1,3)$. The domain of inverse is also this.

Given $f(x)=x^{m^2+2m-8}$ and $m\in \mathbb{Z}$, what values of m can make the function exist at $x=0$?

Since this is a power function and the power will be an integer, negative powers will make asymptotes, meaning that $m^2+2m-8\ge 0$. Solving this, we get $m\in (-\infty ,-4]\cup [2,\infty )$.